(a) Field of the Invention
The present invention relates to a method for calculating frequency offset in an orthogonal frequency division multiplexing system, and more particularly, to a frequency offset calculation method that uses log transforms and linear approximation.
(b) Description of the Related Art
If in orthogonal frequency division multiplexing an offset occurs between transmission and reception frequencies by the imprecision in an oscillator and a difference in carrier wave frequencies of a transmission side and a receiving side, interference between channels results. Also, if this offset accumulates, precise fast Fourier transform (FFT) is unable to be performed on the receiving side to thereby prevent the ability to demodulate data.
Therefore, frequency offset must be obtained by only a short training sequence in an orthogonal frequency division multiplexing system. At this time, frequency offset may be obtained using two or more consecutive and identical training sequence phase differences.
In more detail, if a period of the training sequence Y[n] is N, two training sequence samples received in an interval of N samples are as shown in the following equations.
                                          Y            ⁡                          [              n              ]                                =                                    1              N                        ⁢                                          ∑                                  k                  =                  0                                                  N                  -                  1                                            ⁢                                                          ⁢                                                X                  ⁡                                      [                    k                    ]                                                  ·                                  ⅇ                                      j2π                    ⁢                                                                                  ⁢                                                                  n                        ⁡                                                  (                                                      k                            +                            ɛ0                                                    )                                                                    /                      N                                                                                                          ⁢                                  ⁢                              Y            ⁡                          [                              n                +                N                            ]                                =                                    1              N                        ⁢                                          ∑                                  k                  =                  0                                                  N                  -                  1                                            ⁢                                                          ⁢                                                X                  ⁡                                      [                    k                    ]                                                  ·                                  ⅇ                                      j2π                    ⁢                                                                                  ⁢                                                                  n                        ⁡                                                  (                                                      k                            +                            ɛ1                                                    )                                                                    /                      N                                                                                                                              [                  Equation          ⁢                                          ⁢          1                ]            where X[k] is an Nth carrier wave channel, ε0 is the frequency offset normalized in Y[n], and ε1 is the frequency offset normalized in Y[n+N]. At this time, the phase difference in the received two training sequence samples is calculated using the following equation.2πε=2π(ε0−ε1)=∠{Y[n]*·Y[n+N]}  [Equation 2]where Y[n]* indicates the conjugate value of Y[n], and ∠ is the phase angle calculated using tan−1.
Next, if over the entire length the affect of noise is eliminated by obtaining the integral by as many as N samples, the frequency offset is obtained as follows.
                    ɛ        =                              1                          2              ⁢              π                                ⁢                      ∠            ⁡                          [                                                ∑                                      n                    =                    0                                                        N                    -                    1                                                  ⁢                                  {                                                                                    Y                        ⁡                                                  [                          n                          ]                                                                    *                                        ·                                          Y                      ⁡                                              [                                                  n                          +                          N                                                ]                                                                              }                                            ]                                                          [                  Equation          ⁢                                          ⁢          3                ]            
In the case where the frequency offset is obtained using this method, an approximate frequency offset is found using a short training sequence, which has a short length and is repeated many times, or a precise frequency offset is obtained using a long training sequence, which has a long length and is only repeated two times. The former case is referred to as coarse frequency offset searching, while the latter case is referred to as fine frequency offset searching.
Such a method has been used in conventional orthogonal frequency division multiplexing systems and adopted in standards. An example of this method will be described with reference to FIG. 1.
FIG. 1 is a drawing of a training sequence structure according to the IEEE 802.11a standard. An equation for calculating frequency offset when the two training sequences are Y[n]=I1+jQ1 and Y[n]=I2+jQ2 is as follows.
                    ɛ        =                                            1                              2                ⁢                π                ⁢                                                                  ⁢                N                                      ⁢                          ∠              ⁡                              (                                                                            Y                      ⁡                                              [                        n                        ]                                                              *                                    ·                                      Y                    ⁡                                          [                                              n                        +                        N                                            ]                                                                      )                                              =                                                    1                                  2                  ⁢                  π                  ⁢                                                                          ⁢                  N                                            ⁢                              ∠                ⁡                                  (                                                                                    I                        1                                            ⁢                                              I                        2                                                              +                                                                  Q                        1                                            ⁢                                              Q                        2                                                              +                                          j                      ⁡                                              (                                                                                                            I                              1                                                        ⁢                                                          Q                              2                                                                                -                                                                                    I                              2                                                        ⁢                                                          Q                              1                                                                                                      )                                                                              )                                                      =                                                            1                                      2                    ⁢                    π                    ⁢                                                                                  ⁢                    N                                                  ⁢                                  ∠                  ⁡                                      (                                                                  E                        I                                            +                                              j                        ⁢                                                                                                  ⁢                                                  E                          Q                                                                                      )                                                              =                                                1                                      2                    ⁢                    π                    ⁢                                                                                  ⁢                    N                                                  ⁢                                                      tan                                          -                      1                                                        ⁡                                      (                                                                  E                        Q                                                                    E                        I                                                              )                                                                                                          [                  Equation          ⁢                                          ⁢          4                ]            
An example of logic used to perform this calculation is shown in FIG. 2.
FIG. 2 is a drawing showing a structure of logic used to obtain a frequency offset value in an orthogonal frequency division multiplexing system. When calculating frequency offset using the illustrated logic, it is necessary to use the arc tangent (tan−1), and this is performed using a look-up table as shown in FIG. 3.
As shown in the drawings, the arc tangent look-up table receives the real part E1 and the imaginary part EQ, and outputs the value of
      1          2      ⁢      π      ⁢                          ⁢      N        ⁢                    tan                  -          1                    ⁡              (                              E            Q                                E            I                          )              .  In the case where each of the inputs of the real part E1 and the imaginary part EQ is 4 bits, 16×16=256 bits are needed for the arc tangent look-up table.
Further, if the number of bits of each the real part E1 and the imaginary part EQ is increased to enhance the precision in the resulting value, the size of the look-up table is exponentially increased. This is particularly problematic in fine frequency offset searching in which the difference in the real part and imaginary part values is large and precise calculations are required. For example, if each of the real part and imaginary part is 8 bits, a look-up table of 28×28=65,536 bits is needed.
An increase in the size of the arc tangent table as a result of more bits for the real part and imaginary part of training sequences results in an increase in hardware area and power consumption.
In a related patent, Samsung Electro Mechanics disclosed Korean Patent Publication No. 10-2000-17523 entitled Frequency Offset Precision Compensation Method for High-Speed Wireless LAN. In this patent, efforts are made to enhance the performance of the high-speed wireless LAN by reducing the size of the arc tangent look-up table used to calculate frequency offset. However, a separate, additional logic is required in this patent to reduce the size of the arc tangent look-up table.
Accordingly, there is a need for a simple logic calculation method for the calculation of frequency offset that minimizes hardware area and power consumption.